# Challenging Math, anyone?

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The sum of the first 100 terms of an arithmetic progression is 15050; the first, third and eleventh terms of this progression are three consecutive terms of a geometric progression. Find the first term, a and the non-zero common difference, d, of the arithmetic progression.

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#2

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The sum of the first 100 terms of an arithmetic progression is 15050; the first, third and eleventh terms of this progression are three consecutive terms of a geometric progression. Find the first term, a and the non-zero common difference, d, of the arithmetic progression.

**crystalong**)The sum of the first 100 terms of an arithmetic progression is 15050; the first, third and eleventh terms of this progression are three consecutive terms of a geometric progression. Find the first term, a and the non-zero common difference, d, of the arithmetic progression.

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#3

**crystalong**)

The sum of the first 100 terms of an arithmetic progression is 15050; the first, third and eleventh terms of this progression are three consecutive terms of a geometric progression. Find the first term, a and the non-zero common difference, d, of the arithmetic progression.

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#5

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Haha...how do you solve it?

**crystalong**)Haha...how do you solve it?

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#7

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Both

**crystalong**)Both

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I am pretty sure they know how to do it. Have you put this up because you don't know how or for others to do it?

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**physicsmaths**)I am pretty sure they know how to do it. Have you put this up because you don't know how or for others to do it?

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#10

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No

**crystalong**)No

Also that you do not know the rule for a term

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Are you saying that you do not know the rule for the sum of n terms

Also that you do not know the rule for a term

**TenOfThem**)Are you saying that you do not know the rule for the sum of n terms

Also that you do not know the rule for a term

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#12

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I know the rules but I do not know how to apply it here

**crystalong**)I know the rules but I do not know how to apply it here

For example you can write the third term as a+2d

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So you can write expressions in terms of a and d

For example you can write the third term as a+2d

**TenOfThem**)So you can write expressions in terms of a and d

For example you can write the third term as a+2d

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#14

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No

**crystalong**)No

Sum of an arithmetic series = (n/2)(2a + (n-1)d) With n being the number of terms.

You already have n, so this will form an equation in terms of the unknowns a and d. Now you need to look at the geometric series element to see if another equation can be formed from that.

The first term is simply a. Now think about how you can express the third and eleventh in terms of a and d: with each successive term, you add on an extra d.

But these form a geometric series, with terms a, ar, ar^2 where r is the common ratio. What is ar/a? What is (ar^2)/ar? Each time we move on to another term, we multiply by the same amount. Hopefully this will guide you to another equation in terms of a and d; since you then have two, with only two unknown variables, you can solve for a and d

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#15

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Did that too

**crystalong**)Did that too

You said that you could not write expressions in terms of a and d

Now you suggest that you have done

Do you have an equation in a and d for the sum and three expressions for the terms? Yes or no?

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#16

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Hello! You know how to solve it?

**crystalong**)Hello! You know how to solve it?

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Basically, the key idea here is forming simultaneous equations. I assume you know the basic formula involved but, just in case, it can't hurt to write it down. If you've already done some of the starter steps here I apologise

Sum of an arithmetic series = (n/2)(2a + (n-1)d) With n being the number of terms.

You already have n, so this will form an equation in terms of the unknowns a and d. Now you need to look at the geometric series element to see if another equation can be formed from that.

The first term is simply a. Now think about how you can express the third and eleventh in terms of a and d: with each successive term, you add on an extra d.

But these form a geometric series, with terms a, ar, ar^2 where r is the common ratio. What is ar/a? What is (ar^2)/ar? Each time we move on to another term, we multiply by the same amount. Hopefully this will guide you to another equation in terms of a and d; since you then have two, with only two unknown variables, you can solve for a and d

**13 1 20 8 42**)Basically, the key idea here is forming simultaneous equations. I assume you know the basic formula involved but, just in case, it can't hurt to write it down. If you've already done some of the starter steps here I apologise

Sum of an arithmetic series = (n/2)(2a + (n-1)d) With n being the number of terms.

You already have n, so this will form an equation in terms of the unknowns a and d. Now you need to look at the geometric series element to see if another equation can be formed from that.

The first term is simply a. Now think about how you can express the third and eleventh in terms of a and d: with each successive term, you add on an extra d.

But these form a geometric series, with terms a, ar, ar^2 where r is the common ratio. What is ar/a? What is (ar^2)/ar? Each time we move on to another term, we multiply by the same amount. Hopefully this will guide you to another equation in terms of a and d; since you then have two, with only two unknown variables, you can solve for a and d

50(2a+99d)=15050

2a+99d=301

d=(301-2a)/99

1st term=a

3rd term=a+2d

11st term=a+10d

2((a+2d)/a)=(a+10d)/a

a=6d

=6((301-2a)/99).....

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#18

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a=2,d=3

**brianeverit**)a=2,d=3

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Did you use Smaug's method, which results in three simultaneous equations (one of which involves a square); or is there another way to do it?

**gagafacea1**)Did you use Smaug's method, which results in three simultaneous equations (one of which involves a square); or is there another way to do it?

My workings:

50(2a+99d)=15050

2a+99d=301

d=(301-2a)/99

1st term=a

3rd term=a+2d

11st term=a+10d

2((a+2d)/a)=(a+10d)/a

a=6d

=6((301-2a)/99)

I used this method and got a and d with decimals

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#20

(Original post by

My workings:

50(2a+99d)=15050

2a+99d=301

d=(301-2a)/99

1st term=a

3rd term=a+2d

11st term=a+10d

2((a+2d)/a)=(a+10d)/a

a=6d

=6((301-2a)/99).....

**crystalong**)My workings:

50(2a+99d)=15050

2a+99d=301

d=(301-2a)/99

1st term=a

3rd term=a+2d

11st term=a+10d

2((a+2d)/a)=(a+10d)/a

a=6d

=6((301-2a)/99).....

I don't know where you got 2((a + 2d)/a) = (a + 10d)/a from though

Considering the common ratio to be r

a = a

a + 2d = ar

a + 10d = ar^2

Your equation says that 2r = r^2; what you want to do is get something that says r = r, but where r is in terms of a and d.

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